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emcrazy14 Group Title

Differentiate; x^2/(1+x)^2

  • one year ago
  • one year ago

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  1. ajprincess Group Title
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    This is of the form u/v. (u/v)'=(u'v-v'u)/v^2 Here u=x^2 and v=(1+x)^2. (x^n)'=nx^(n-1) Can u do it @emcrazy14?

    • one year ago
  2. emcrazy14 Group Title
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    I got the answer -2x/(1+x)^3. Is it correct?

    • one year ago
  3. experimentX Group Title
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    differentiate: 1/(1/x + 1)^2

    • one year ago
  4. ajprincess Group Title
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    check the sign @emcrazy14. It shud be 2x/(1+x)^3

    • one year ago
  5. emcrazy14 Group Title
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    Ok, thanks. Do I multiply (x+1) with a negative sign to change it into (1+x) so that it cancels with the the denominator (1+x)^4?

    • one year ago
  6. experimentX Group Title
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    just ask the wolf for differentiation Question.

    • one year ago
  7. ajprincess Group Title
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    nope. (x+1)=(1+x)

    • one year ago
  8. emcrazy14 Group Title
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    Then how oes it become +2x? :O

    • one year ago
  9. emcrazy14 Group Title
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    @experimentX : wolf? :O *does

    • one year ago
  10. experimentX Group Title
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    http://www.wolframalpha.com/input/?i=differentiate+x^2%2F%281%2Bx%29^2

    • one year ago
  11. ajprincess Group Title
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    |dw:1352469822062:dw||dw:1352469940166:dw||dw:1352470079425:dw|

    • one year ago
  12. experimentX Group Title
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    \[ x^2/(1+x)^2 = 1/(1/x + 1)^2 = ({1 \over x} + 1)^{-2}\] use chain rule ... this should be shorter.

    • one year ago
  13. emcrazy14 Group Title
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    @ajprincess : Get it now. Thank you so much! :D

    • one year ago
  14. ajprincess Group Title
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    welcome:) @experimentX's method is much more simpler. U can use that too.:)

    • one year ago
  15. emcrazy14 Group Title
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    @experimentX : Once I perfect this one, I'll use the one you mentioned. Thank you! :)

    • one year ago
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